Multiply the following complex numbers: $({5+4i}) \cdot ({4-2i})$
Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({5+4i}) \cdot ({4-2i}) = $ $ ({5} \cdot {4}) + ({5} \cdot {-2}i) + ({4}i \cdot {4}) + ({4}i \cdot {-2}i) $ Then simplify the terms: $ (20) + (-10i) + (16i) + (-8 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ 20 + (-10 + 16)i - 8i^2 $ After we plug in $i^2 = -1$ , the result becomes $ 20 + (-10 + 16)i - (-8) $ The result is simplified: $ (20 + 8) + (6i) = 28+6i $